Problem 16 - A Complete Resource Book in Chemistry for JEE Main

Question

The equivalent weight of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) in the reaction \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\) is \((\mathrm{M}=\) molecular weight \()\) (a) \(\mathrm{M}\) (b) \(\mathrm{M} / 4\) (c) \(\mathrm{M} / 0.5\) (d) \(\mathrm{M} / 2\)

Step-by-Step Solution

Answer

The equivalent weight is \(\frac{M}{2}\), so the answer is (d).

Step 1: Understanding the Concept of Equivalent Weight

The equivalent weight of a compound is the mass that releases or consumes one mole of valence electrons in a reaction. In other words, it's the molecular weight divided by the number of electrons lost or gained per molecule of the compound in the reaction.

Step 2: Analyze the Reaction

The given reaction is \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} + \mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} + 2 \mathrm{NaI}\). Each \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) unit passes one electron to \(\mathrm{I}_{2}\), allowing two \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) units together to pass two electrons for the reaction to occur.

Step 3: Calculate the Equivalent Weight

For the equivalent weight, consider how many moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are needed to transfer one mole of electrons. Since 2 moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are needed to transfer 2 moles of electrons, the equivalent weight of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is \(\frac{\text{Molecular Weight (M)}}{2}\).

Step 4: Final Calculation and Conclusion

Thus the equivalent weight of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) in the reaction is \(\frac{M}{2}\). Hence, the correct option is (d).

Key Concepts

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. These electrons are crucial in determining how an atom will interact or bond with other atoms. When dealing with chemical reactions, especially ones involving redox processes, valence electrons play a pivotal role.
In the context of the equivalent weight calculation:

• It’s the valence electrons that are gained or lost in reactions, indicating the reactivity of a molecule.
• Understanding which electrons are valence can help predict the product of the chemical reaction.

As in the given reaction, where iodide accepts electrons passed by sodium thiosulfate, understanding valence electrons helps in calculating how many moles are necessary to achieve the charge balance. This insight is essential for calculating the equivalent weight of compounds in such reactions.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, convert into different substances, called products. There are various types of chemical reactions, such as synthesis, decomposition, single replacement, and the type presented here, which is a redox reaction.
In the example reaction:

• The transformation involves sodium thiosulfate and iodine, producing sodium tetrathionate and sodium iodide.
• A key aspect of this transformation is the electron transfer that takes place.

This is a redox reaction, where oxidation and reduction occur simultaneously, making the observation of electron transfer from one reactant to another crucial for understanding how the reaction proceeds and for subsequent calculations like equivalent weight determination.

Molecular Weight

Molecular weight, also known as molecular mass, is the sum of the atomic weights of all atoms in a molecule. It's usually expressed in atomic mass units (amu) or grams per mole.
For the purpose of calculating equivalent weight:

• The molecular weight is a critical value as it's divided by the number of electrons transferred to find the equivalent weight.
• This weight provides the baseline measure for converting the moles of substance involved in a reaction into available electrons for other reactants in the equation.

Precisely calculating the molecular weight of compounds like sodium thiosulfate in the given chemical equation is crucial to achieving the accurate equivalent weight.

Redox Reaction Analysis

Redox (reduction-oxidation) reactions involve the transfer of electrons from one molecule to another and are fundamental concepts in chemistry. In such reactions:

• Oxidation refers to the loss of electrons. In the reaction, sodium thiosulfate loses electrons, performing the function of a reducing agent.
• Reduction is the gain of electrons; here, iodine gains electrons, illustrating its role as an oxidizing agent.

Analyzing these reactions helps understand the electron flow, which is crucial for equivalent weight calculations. For sodium thiosulfate, recognizing that two of its molecules lose electrons to one molecule of iodine is key for determining how the equivalent weight is derived. This understanding ensures that chemical balance in reactions is maintained.